wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s641351127
|
p03455
|
u375681664
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 269 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
S = input()
if S in "1234567890":
print ("number")
elif S in "abcdefghijklmnopqrstuvwxyz":
print ("alphabet")
else:
print ("other")
|
s830931134
|
Accepted
| 17 | 2,940 | 127 |
a,b = input().split()
a = int(a)
b = int(b)
c = a % 2
d = b % 2
if c == 0 or d == 0:
print ("Even")
else:
print ("Odd")
|
s744391986
|
p03090
|
u189575640
| 2,000 | 1,048,576 |
Wrong Answer
| 295 | 21,212 | 1,037 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
from sys import exit
from numpy.random import rand
# A,B,C = [int(n) for n in input().split()]
# N = int(input())
# a = [int(input()) for _ in range(N)]
# S = str(input())
# L = len(S)
# T = str(input())
N = int(input())
# dic = {}
ansgroup = []
if N%2 == 0:
for i in range(1,N//2+1):
ansgroup.append([i, N-i+1])
else:
for i in range(1,N//2+1):
ansgroup.append([i, N-i])
ansgroup.append([N])
# print(ansgroup)
for i in range(len(ansgroup)):
for j in range(i+1,len(ansgroup)):
for e1 in ansgroup[i]:
for e2 in ansgroup[j]:
print(e1,e2)
# tmpset = set()
# if ans[i-1][0] == ans[i][0]:
# else:
# exit()
|
s004161132
|
Accepted
| 158 | 14,512 | 1,124 |
from sys import exit
from numpy.random import rand
# A,B,C = [int(n) for n in input().split()]
# N = int(input())
# a = [int(input()) for _ in range(N)]
# S = str(input())
# L = len(S)
# T = str(input())
N = int(input())
# dic = {}
ansgroup = []
if N%2 == 0:
for i in range(1,N//2+1):
ansgroup.append([i, N-i+1])
else:
for i in range(1,N//2+1):
ansgroup.append([i, N-i])
ansgroup.append([N])
nodesum = 0
ans = []
for i in range(len(ansgroup)):
for j in range(i+1,len(ansgroup)):
for e1 in ansgroup[i]:
for e2 in ansgroup[j]:
nodesum+=1
ans.append([e1,e2])
print(nodesum)
for a in ans:
print(a[0],a[1])
# tmpset = set()
# if ans[i-1][0] == ans[i][0]:
# else:
# exit()
|
s753094690
|
p02394
|
u446066125
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 149 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
(W, H, x, y, r) = [int(i) for i in input().split()]
if 0 >= x - r and 0 >= y - r and W <= x + r and H <= y + r :
print('Yes')
else :
print('No')
|
s763373627
|
Accepted
| 40 | 6,724 | 149 |
(W,H,x,y,r) = [int(i) for i in input().split()]
if x - r >= 0 and y - r >= 0 and x + r <= W and y + r <= H :
print('Yes')
else :
print('No')
|
s393217804
|
p03478
|
u350997995
| 2,000 | 262,144 |
Wrong Answer
| 34 | 3,060 | 155 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N,A,B = map(int,input().split())
ans = 0
for i in range(1,N+1):
S = str(i)
a = 0
for s in S:
a+=int(s)
if A<=a<=B:ans+=1
print(ans)
|
s716467833
|
Accepted
| 34 | 3,060 | 155 |
N,A,B = map(int,input().split())
ans = 0
for i in range(1,N+1):
S = str(i)
a = 0
for s in S:
a+=int(s)
if A<=a<=B:ans+=i
print(ans)
|
s712044895
|
p03680
|
u722535636
| 2,000 | 262,144 |
Wrong Answer
| 175 | 7,852 | 193 |
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
n = int(input())
l = [0] + [int(input()) for i in range(n)]
cnt = 0
a = 1
while a != 2:
if a==0:
print(-1)
break
a = l[a]
l[a] = 0
cnt += 1
else:
print(cnt)
|
s881281843
|
Accepted
| 205 | 7,852 | 201 |
n = int(input())
l = [0] + [int(input()) for i in range(n)]
cnt = 0
a = 1
while a != 2:
if a==0:
print(-1)
break
tmp=l[a]
l[a]=0
a=tmp
cnt += 1
else:
print(cnt)
|
s663231490
|
p03711
|
u746849814
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 221 |
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int, input().split())
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
if x in a and y in b:
print('Yes')
elif x in b and y in b:
print('Yes')
elif x in c and y in c:
print('Yes')
else:
print('No')
|
s058005879
|
Accepted
| 17 | 3,060 | 221 |
x, y = map(int, input().split())
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
if x in a and y in a:
print('Yes')
elif x in b and y in b:
print('Yes')
elif x in c and y in c:
print('Yes')
else:
print('No')
|
s840821258
|
p02316
|
u947762778
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,636 | 277 |
You have N kinds of items that you want to put them into a knapsack. Item i has value vi and weight wi. You want to find a subset of items to put such that: * The total value of the items is as large as possible. * The items have combined weight at most W, that is capacity of the knapsack. * You can select as many items as possible into a knapsack for each kind. Find the maximum total value of items in the knapsack.
|
N, W = map(int, input().split())
value = []
weight = []
dp = [0 for _ in range(W + 1)]
for i in range(N):
v, w = map(int, input().split())
for j in range(w+1):
if j >= w:
dp[j] = max(dp[j], dp[j - w] + v)
else:
break
print(dp[W])
|
s592927724
|
Accepted
| 680 | 8,024 | 245 |
N, W = map(int, input().split())
value = []
weight = []
dp = [0 for _ in range(W + 1)]
for i in range(N):
v, w = map(int, input().split())
for j in range(W+1):
if j >= w:
dp[j] = max(dp[j], dp[j - w] + v)
print(dp[W])
|
s652689491
|
p03657
|
u089032001
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 102 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
print("Possible" if a+b%3 == 0 or a%3==0 or b%3==0 else "Impossible")
|
s576786709
|
Accepted
| 18 | 2,940 | 104 |
a, b = map(int, input().split())
print("Possible" if (a+b)%3 == 0 or a%3==0 or b%3==0 else "Impossible")
|
s165098219
|
p02399
|
u442346200
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,744 | 105 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
(a, b) = [int(x) for x in input().split()]
x = a // b
y = a % b
z = a / b
print(x);
print(y);
print(z);
|
s944125979
|
Accepted
| 30 | 6,748 | 110 |
(a, b) = [int(i) for i in input().split()]
d = int (a / b)
r = a % b
f = a / b
print('%s %s %.5f' % (d, r, f))
|
s076732042
|
p03069
|
u101742684
| 2,000 | 1,048,576 |
Wrong Answer
| 957 | 23,844 | 390 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
import numpy as np
N = int(input())
s = input()
S = []
for c in s:
S.append(c)
A = np.zeros(N+1)
if S[0] == "#":
A[0] = 0
for i in range(1,N+1):
if S[i-1] == "#":
A[i] = A[i-1] + 1
else:
A[i] = A[i-1]
minimum = N
for i in range(N+1):
num = int(A[i] + (N-i-(A[N]-A[i])))
if minimum > num:
minimum = num
print(num)
|
s732171672
|
Accepted
| 949 | 18,132 | 393 |
import numpy as np
N = int(input())
s = input()
S = []
for c in s:
S.append(c)
A = np.zeros(N+1)
if S[0] == "#":
A[0] = 0
for i in range(1,N+1):
if S[i-1] == "#":
A[i] = A[i-1] + 1
else:
A[i] = A[i-1]
minimum = N
for i in range(N+1):
num = int(A[i] + (N-i-(A[N]-A[i])))
if minimum > num:
minimum = num
print(minimum)
|
s330932904
|
p02392
|
u801346721
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,496 | 110 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a, b, c = input().split(' ')
a = int(a)
b = int(b)
c = int(c)
if a > b > c:
print('Yes')
else:
print('No')
|
s927771038
|
Accepted
| 60 | 7,648 | 110 |
a, b, c = input().split(' ')
a = int(a)
b = int(b)
c = int(c)
if a < b < c:
print("Yes")
else:
print("No")
|
s575705185
|
p03476
|
u666964944
| 2,000 | 262,144 |
Wrong Answer
| 603 | 12,180 | 241 |
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
q = int(input())
n = 100001
limit = int(n**0.5)
a = [i for i in range(3, n, 2)]
e = []
for i in range(n):
p = a[i]
if limit <= p:
break
e = [j for j in a if j%p!=0 ]
print(e)
for i in range(q):
l,r = map(int, input().split())
|
s936270358
|
Accepted
| 1,465 | 14,424 | 670 |
#import numpy
#for loop
#a = [1,2,3,4,5,6,7,8,9,10]
#s = [0]*len(a)
#s[0] = a[0]
# s[i+1] = s[i] + a[i+1]
#print(s)
#numpy.cumsum()
#print(numpy.cumsum(a))
q = int(input())
n = 10**5+1
limit = int(n**0.5)
a = [i for i in range(3, n, 2)]
prime = [2]
while True:
p = a[0]
if limit <= p:
prime += a
break
prime.append(p)
a = [j for j in a if j%p!=0]
pi = [0]*n
for i in range(len(prime)):
ai = (prime[i]+1)//2
if ai in prime:
pi[prime[i]-1] = 1
s = [0]*n
s[0] = pi[0]
for i in range(len(pi)-1):
s[i+1] = s[i] + pi[i+1]
s = [0,0] + s
for i in range(q):
l, r = map(int, input().split())
print(s[r+1]-s[l])
|
s502618961
|
p03339
|
u287775083
| 2,000 | 1,048,576 |
Wrong Answer
| 133 | 15,140 | 969 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
import sys
import re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
N = INT()
S = [_ for _ in input()]
init_cnt = S[1:].count('E')
min_c = init_cnt
ans = 0
for i in range(1, N):
tmp = min_c
if S[i-1] == 'W':
tmp = tmp + 1
if S[i] == 'E':
tmp = tmp - 1
if tmp < min_c:
min_c = tmp
ans = i
print(ans)
|
s488212969
|
Accepted
| 171 | 15,136 | 923 |
import sys
import re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, gcd
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
N = INT()
S = [_ for _ in input()]
left, right = 0, S[1:].count('E')
ans = left + right
for i in range(N-1):
if S[i] == 'W':
left += 1
if S[i+1] == 'E':
right -= 1
ans = min(ans, left + right)
print(ans)
|
s183741746
|
p03578
|
u692632484
| 2,000 | 262,144 |
Wrong Answer
| 390 | 35,556 | 299 |
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
N=int(input())
D=[int(i) for i in input().split()]
M=int(input())
T=[int(i) for i in input().split()]
D.sort()
T.sort()
indexD=0
indexT=0
while indexD<len(D) and indexT<len(T):
if D[indexD]==T[indexT]:
indexD+=1
indexT+=1
else:
indexD+=1
if indexT==len(T):
print("Yes")
else:
print("No")
|
s467225998
|
Accepted
| 426 | 35,556 | 299 |
N=int(input())
D=[int(i) for i in input().split()]
M=int(input())
T=[int(i) for i in input().split()]
D.sort()
T.sort()
indexD=0
indexT=0
while indexD<len(D) and indexT<len(T):
if D[indexD]==T[indexT]:
indexD+=1
indexT+=1
else:
indexD+=1
if indexT==len(T):
print("YES")
else:
print("NO")
|
s784381160
|
p00004
|
u811733736
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,732 | 531 |
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
import sys
from math import fabs
if __name__ == '__main__':
epsilon = 1e-10
for line in sys.stdin:
a, b, e, c, d, f = [int(x) for x in line.strip().split(' ')]
x = 1/(a*d-b*c) * (d*e - b*f)
y = 1/(a*d-b*c) * (-c*e + a*f)
if fabs(x) < epsilon:
x = 0.0
if fabs(y) < epsilon:
y = 0.0
print('{0:.3f}, {1:.3f}'.format(x, y))
|
s683585649
|
Accepted
| 30 | 7,780 | 528 |
import sys
from math import fabs
if __name__ == '__main__':
epsilon = 1e-9
for line in sys.stdin:
a, b, e, c, d, f = [int(x) for x in line.strip().split(' ')]
x = 1/(a*d-b*c) * (d*e - b*f)
y = 1/(a*d-b*c) * (-c*e + a*f)
if fabs(x) < epsilon:
x = 0.0
if fabs(y) < epsilon:
y = 0.0
print('{0:.3f} {1:.3f}'.format(x, y))
|
s798537636
|
p03494
|
u360258922
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 184 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = input()
a = list(map(int, input().split(' ')))
cnt = 0
while sum(x%2==0 for x in a) == n:
a = list(map(lambda x: x / 2 if x % 2 == 0 else x, a))
cnt += 1
print(cnt)
|
s989110393
|
Accepted
| 20 | 3,060 | 181 |
n = int(input())
a = list(map(int, input().split(' ')))
cnt = 0
while sum(x%2==0 for x in a) == n:
a = list(map(lambda x: x / 2 if x % 2 == 0 else x, a))
cnt += 1
print(cnt)
|
s285644005
|
p02927
|
u629350026
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,020 | 147 |
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
m,d=map(int,input().split())
ans=0
for i in range(1,m+1):
for j in range(22,d+1):
if j%10>=2 and (j/10)*(j%10)==i:
ans=ans+1
print(ans)
|
s405207454
|
Accepted
| 26 | 9,084 | 148 |
m,d=map(int,input().split())
ans=0
for i in range(1,m+1):
for j in range(22,d+1):
if j%10>=2 and (j//10)*(j%10)==i:
ans=ans+1
print(ans)
|
s726888085
|
p03494
|
u064408584
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 137 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
a=list(map(int, input().split()))
count=0
while sum(a)%2==0:
a=[i/2 for i in a]
count+=1
print(a)
print(count)
|
s687924141
|
Accepted
| 19 | 3,060 | 136 |
n=int(input())
a=list(map(int, input().split()))
count=0
while all(i%2==0 for i in a):
a=[i//2 for i in a]
count+=1
print(count)
|
s638959156
|
p02669
|
u162612857
| 2,000 | 1,048,576 |
Wrong Answer
| 408 | 11,016 | 1,802 |
You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.**
|
import math
def calc_cost(n, ans_dict):
min_cost = n * d
inc_2 = int(math.ceil(n / 2))
if inc_2 not in ans_dict:
ans_dict[inc_2] = calc_cost(inc_2, ans_dict)
min_cost = min(min_cost, d * (inc_2 * 2 - n) + a + ans_dict[inc_2])
dec_2 = int(math.floor(n / 2))
if dec_2 not in ans_dict:
ans_dict[dec_2] = calc_cost(dec_2, ans_dict)
min_cost = min(min_cost, d * (- dec_2 * 2 + n) + a + ans_dict[dec_2])
inc_3 = int(math.ceil(n / 3))
if inc_3 not in ans_dict:
ans_dict[inc_3] = calc_cost(inc_3, ans_dict)
min_cost = min(min_cost, d * (inc_3 * 3 - n) + b + ans_dict[inc_3])
dec_3 = int(math.floor(n / 3))
if dec_3 not in ans_dict:
ans_dict[dec_3] = calc_cost(dec_3, ans_dict)
min_cost = min(min_cost, d * (- dec_3 * 3 + n) + b + ans_dict[dec_3])
inc_5 = int(math.ceil(n / 5))
if inc_5 not in ans_dict:
ans_dict[inc_5] = calc_cost(inc_5, ans_dict)
min_cost = min(min_cost, d * (inc_5 * 5 - n) + c + ans_dict[inc_5])
dec_5 = int(math.floor(n / 5))
if dec_5 not in ans_dict:
ans_dict[dec_5] = calc_cost(dec_5, ans_dict)
min_cost = min(min_cost, d * (- dec_5 * 5 + n) + c + ans_dict[dec_5])
ans_dict[n] = min_cost
return min_cost
n_testcase = int(input())
for i in range(n_testcase):
n, a, b, c, d = list(map(int, input().split()))
ans_dict = {}
ans_dict[0] = 0
ans_dict[1] = d
ans = calc_cost(n, ans_dict)
print(ans)
# print(ans_dict)
|
s475117787
|
Accepted
| 312 | 11,124 | 1,726 |
def calc_cost(n, ans_dict):
min_cost = n * d
inc_2 = (n + 2 - 1) // 2
if inc_2 not in ans_dict:
ans_dict[inc_2] = calc_cost(inc_2, ans_dict)
min_cost = min(min_cost, d * (inc_2 * 2 - n) + a + ans_dict[inc_2])
dec_2 = n // 2
if dec_2 not in ans_dict:
ans_dict[dec_2] = calc_cost(dec_2, ans_dict)
min_cost = min(min_cost, d * (- dec_2 * 2 + n) + a + ans_dict[dec_2])
inc_3 = (n + 3 - 1) // 3
if inc_3 not in ans_dict:
ans_dict[inc_3] = calc_cost(inc_3, ans_dict)
min_cost = min(min_cost, d * (inc_3 * 3 - n) + b + ans_dict[inc_3])
dec_3 = n // 3
if dec_3 not in ans_dict:
ans_dict[dec_3] = calc_cost(dec_3, ans_dict)
min_cost = min(min_cost, d * (- dec_3 * 3 + n) + b + ans_dict[dec_3])
inc_5 = (n + 5 - 1) // 5
if inc_5 not in ans_dict:
ans_dict[inc_5] = calc_cost(inc_5, ans_dict)
min_cost = min(min_cost, d * (inc_5 * 5 - n) + c + ans_dict[inc_5])
dec_5 = n // 5
if dec_5 not in ans_dict:
ans_dict[dec_5] = calc_cost(dec_5, ans_dict)
min_cost = min(min_cost, d * (- dec_5 * 5 + n) + c + ans_dict[dec_5])
ans_dict[n] = min_cost
return min_cost
n_testcase = int(input())
for i in range(n_testcase):
n, a, b, c, d = list(map(int, input().split()))
ans_dict = {}
ans_dict[0] = 0
ans_dict[1] = d
ans = calc_cost(n, ans_dict)
print(ans)
# print(ans_dict)
|
s341126813
|
p03478
|
u813569174
| 2,000 | 262,144 |
Wrong Answer
| 30 | 3,060 | 196 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
s = input().split()
N = int(s[0])
A = int(s[1])
B = int(s[2])
k = 0
for i in range(1,N+1):
S = 0
while i >= 1:
p = i%10
i = i//10
S = S + p
if A <= S <= B:
k = k + i
print(k)
|
s357607127
|
Accepted
| 30 | 3,060 | 204 |
s = input().split()
N = int(s[0])
A = int(s[1])
B = int(s[2])
k = 0
for i in range(1,N+1):
S = 0
l = i
while l >= 1:
p = l%10
l = l//10
S = S + p
if A <= S <= B:
k = k + i
print(k)
|
s243262250
|
p03149
|
u849756457
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,068 | 137 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n = [int(i) for i in input().split()]
n.sort()
if n[0] == 1 and n[1] == 4 and n[2] == 7 and n[3] == 9:
print("Yes")
else:
print("No")
|
s689845398
|
Accepted
| 30 | 9,024 | 138 |
n = [int(i) for i in input().split()]
n.sort()
if n[0] == 1 and n[1] == 4 and n[2] == 7 and n[3] == 9:
print("YES")
else:
print("NO")
|
s005846152
|
p02255
|
u567380442
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 205 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n = input()
A = list(map(int, input().split()))
for i in range(1, len(A)):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j+1] = A[j]
j -= 1
A[j+1] = key
print(*A)
|
s222528180
|
Accepted
| 40 | 6,756 | 214 |
n = input()
A = list(map(int, input().split()))
print(*A)
for i in range(1, len(A)):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j+1] = A[j]
j -= 1
A[j+1] = key
print(*A)
|
s835235700
|
p02842
|
u256027816
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 255 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n = int(input())
print(math.floor(math.floor(n/1.08)*1.08))
if math.floor(math.floor(n/1.08)*1.08) == n:
print(math.floor(n/1.08))
elif math.floor((math.floor(n/1.08)+1) *1.08) == n:
print(math.floor(n/1.08)+1)
else:
print(':(')
|
s467511933
|
Accepted
| 18 | 3,060 | 212 |
import math
n = int(input())
if math.floor(math.floor(n/1.08)*1.08) == n:
print(math.floor(n/1.08))
elif math.floor((math.floor(n/1.08)+1) *1.08) == n:
print(math.floor(n/1.08)+1)
else:
print(':(')
|
s326344805
|
p03435
|
u811176339
| 2,000 | 262,144 |
Wrong Answer
| 35 | 9,228 | 643 |
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
lc = [[int(w) for w in input().split()] for i in range(3)]
def ch1(la, lb, aindex):
return la[aindex] + lb[0] == lc[aindex][0]\
and la[aindex] + lb[1] == lc[aindex][1]\
and la[aindex] + lb[2] == lc[aindex][2]
cond = False
for i in range(100):
if cond:
break
for j in range(100):
la = [i//100, (i % 100)//10, i % 10]
lb = [j//100, (j % 100)//10, j % 10]
if ch1(la, lb, 0):
if ch1(la, lb, 1):
if ch1(la, lb, 2):
cond = True
else:
j += 9
else:
j += 99
print("Yes" if cond else "No")
|
s136881350
|
Accepted
| 30 | 9,132 | 293 |
lc = [[int(w) for w in input().split()] for i in range(3)]
cond = True
lb = lc[0]
a2 = lc[1][0]-lb[0]
a3 = lc[2][0]-lb[0]
la = [0, a2, a3]
for i in range(3):
for j in range(3):
if lc[i][j] != la[i]+lb[j]:
cond = False
break
print("Yes" if cond else "No")
|
s717736507
|
p03477
|
u778700306
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 149 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d = map(int, input().split())
if a + b < c + d:
print("Left")
if a + b > c + d:
print("Right")
if a + b == c + d:
print("Balance")
|
s672260808
|
Accepted
| 18 | 2,940 | 150 |
a,b,c,d = map(int, input().split())
if a + b > c + d:
print("Left")
if a + b < c + d:
print("Right")
if a + b == c + d:
print("Balanced")
|
s221307427
|
p03471
|
u937642029
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 3,064 | 715 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
y,n1 = map(int,input().split())
k10=-1
k5=-1
k1=-1
n=n1
for i in range(n//10000+1):
n=n1
if(n//10000==y and n%10000==0):
k10=n//10000
k5=0
k1=0
break
else:
i=n//10000-i
n-=i*10000
for j in range(n//5000+1):
n=n1
if(n//5000==(y-i) and n%5000==0):
k10=i
k5=n//5000
k1=0
break
else:
j=n//5000-j
n-=j*5000
if(n//1000==(y-i-j) and n%1000==0):
k10=i
k5=j
k1=n//1000
break
if(k1!=-1):
break
print(k10,k5,k1)
|
s314860723
|
Accepted
| 56 | 3,064 | 839 |
y,n1 = map(int,input().split())
k10=-1
k5=-1
k1=-1
n=n1
for i in range(n//10000+1):
n=n1
if(n//10000==y and n%10000==0):
k10=n//10000
k5=0
k1=0
break
elif(n//10000>y):
break
else:
i=n//10000-i
n-=i*10000
for j in range(n//5000+1):
n=n1-i*10000
if(n//5000==(y-i) and n%5000==0):
k10=i
k5=n//5000
k1=0
break
elif(n//5000>y-i):
break
else:
n=n1-i*10000
j=n//5000-j
n-=j*5000
if(n//1000==(y-i-j) and n%1000==0):
k10=i
k5=j
k1=n//1000
break
if(k1!=-1):
break
print(k10,k5,k1)
|
s556904942
|
p04043
|
u077291787
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 185 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = input().rstrip().split()
print("YES" if len(a) == 5 and len(b) == 7 and len(c) == 5 else "NO")
|
s361678818
|
Accepted
| 17 | 2,940 | 233 |
def main():
A = sorted(map(int, input().split()))
flg = A == [5, 5, 7]
print("YES" if flg else "NO")
if __name__ == "__main__":
main()
|
s740672605
|
p02841
|
u138781768
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 110 |
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
if m1 != m2:
print(0)
else:
print(1)
|
s616323421
|
Accepted
| 17 | 2,940 | 110 |
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
if m1 != m2:
print(1)
else:
print(0)
|
s620718092
|
p02612
|
u405733072
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,092 | 71 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
pay = 0
while N <= pay:
pay += 1000
print(pay -N)
|
s155316629
|
Accepted
| 28 | 9,152 | 66 |
N = int(input())
pay = 0
while N > pay:
pay += 1000
print(pay-N)
|
s790109427
|
p03997
|
u597017430
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s160892964
|
Accepted
| 17 | 2,940 | 69 |
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s419872463
|
p03673
|
u190405389
| 2,000 | 262,144 |
Wrong Answer
| 100 | 28,956 | 162 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
a = input().split()
b = [0]*n
for i in range(n//2):
b[i] = a[-1-2*i]
b[n-i-1] = a[-2-2*i]
if n % 2 == 1:
b[n//2] = a[0]
print(b)
|
s295435301
|
Accepted
| 81 | 25,540 | 172 |
n = int(input())
a = input().split()
b = [0]*n
for i in range(n//2):
b[i] = a[-1-2*i]
b[n-i-1] = a[-2-2*i]
if n % 2 == 1:
b[n//2] = a[0]
print(' '.join(b))
|
s525763278
|
p02936
|
u712719923
| 2,000 | 1,048,576 |
Wrong Answer
| 2,207 | 58,908 | 622 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
from collections import deque
def main():
n, q = list(map(int, input().split(" ")))
v_dict = {i: [] for i in range(n)}
for _ in range(n-1):
a, b = list(map(int, input().split(" ")))
v_dict[a-1].append(b-1)
point_list = [0]*n
for _ in range(q):
p, x = list(map(int, input().split(" ")))
stack = deque([p-1])
while stack:
cur = stack.pop()
point_list[cur]+=x
for next_i in v_dict[cur]:
stack.append(next_i)
for i in range(n):
print(point_list[i], end=" ")
if __name__=="__main__":
main()
|
s399241466
|
Accepted
| 1,151 | 62,864 | 811 |
from collections import deque
def main():
n, q = list(map(int, input().split(" ")))
v_list = [[] for i in range(n)]
for _ in range(n-1):
a, b = list(map(int, input().split(" ")))
v_list[a-1].append(b-1)
v_list[b-1].append(a-1)
p_x_list = [0]*n
for _ in range(q):
p, x = list(map(int, input().split(" ")))
p_x_list[p-1]+=x
point_list = [0]*n
checked_list = [0]*n
stack = deque([0])
while stack:
cur = stack.pop()
checked_list[cur] = 1
point_list[cur]+=p_x_list[cur]
for next_i in v_list[cur]:
if checked_list[next_i]:
continue
stack.append(next_i)
point_list[next_i]+=point_list[cur]
print(*point_list)
if __name__=="__main__":
main()
|
s611203138
|
p03943
|
u123273712
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,444 | 227 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
import copy
a = list(map(int,input().split()))
nokori = copy.copy(a)
count = max(a)
nokori.remove(count)
if a[0] == a[1] == a[2]:
print("NO")
if int(nokori[0] + nokori[1]) == count:
print("YES")
else:
print("NO")
|
s124525551
|
Accepted
| 22 | 3,444 | 235 |
import copy
a = list(map(int,input().split()))
nokori = copy.copy(a)
count = max(a)
nokori.remove(count)
"""
if a[0] == a[1] == a[2]:
print("No")
"""
if int(nokori[0] + nokori[1]) == count:
print("Yes")
else:
print("No")
|
s618424707
|
p03815
|
u787562674
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 153 |
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.
|
x = int(input())
if x <= 6:
print(1)
else:
div = x % 11
if div <= 6:
print(2 * x // 11 + 1)
else:
print(2 * x // 11 + 2)
|
s743594609
|
Accepted
| 17 | 3,060 | 241 |
x = int(input())
if x <= 6:
print(1)
elif 7 <= x <= 10:
print(2)
else:
div = x % 11
shou = x // 11
if div == 0:
print(2 * shou)
elif div <= 6:
print(2 * shou + 1)
else:
print(2 * shou + 2)
|
s726922442
|
p01101
|
u843517396
| 8,000 | 262,144 |
Wrong Answer
| 1,570 | 5,636 | 452 |
Mammy decided to give Taro his first shopping experience. Mammy tells him to choose any two items he wants from those listed in the shopping catalogue, but Taro cannot decide which two, as all the items look attractive. Thus he plans to buy the pair of two items with the highest price sum, not exceeding the amount Mammy allows. As getting two of the same item is boring, he wants two different items. You are asked to help Taro select the two items. The price list for all of the items is given. Among pairs of two items in the list, find the pair with the highest price sum not exceeding the allowed amount, and report the sum. Taro is buying two items, not one, nor three, nor more. Note that, two or more items in the list may be priced equally.
|
while 1:
str=input().split()
n=int(str[0])
m=int(str[1])
if n==0 and m==0: break
str=input().split()
a=[0]*n
for i in range(n):
a[i]=int(str[i])
a.sort()
a.reverse()
print(a)
flag=0
max=0
for i in range(n):
if flag!=0: break
for j in range(i+1,n):
v=a[i]+a[j]
if m>=v:
if max<v: max=v
if max==0: print("NONE")
else: print(max)
|
s698146496
|
Accepted
| 1,570 | 5,624 | 439 |
while 1:
str=input().split()
n=int(str[0])
m=int(str[1])
if n==0 and m==0: break
str=input().split()
a=[0]*n
for i in range(n):
a[i]=int(str[i])
a.sort()
a.reverse()
flag=0
max=0
for i in range(n):
if flag!=0: break
for j in range(i+1,n):
v=a[i]+a[j]
if m>=v:
if max<v: max=v
if max==0: print("NONE")
else: print(max)
|
s242953809
|
p02613
|
u853728588
| 2,000 | 1,048,576 |
Wrong Answer
| 138 | 16,152 | 286 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
s = [input() for i in range(N)]
ac = s.count("AC")
wa = s.count("wa")
tle = s.count("TLE")
re = s.count("RE")
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s485336000
|
Accepted
| 139 | 16,276 | 286 |
N = int(input())
s = [input() for i in range(N)]
ac = s.count("AC")
wa = s.count("WA")
tle = s.count("TLE")
re = s.count("RE")
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s942285147
|
p04035
|
u543954314
| 2,000 | 262,144 |
Wrong Answer
| 65 | 14,052 | 254 |
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
|
n,l = map(int,input().split())
a = list(map(int,input().split()))
for i in range(n-1):
if a[i] + a[i+1] >= l:
print("Possible")
break
else:
print("Impossible")
exit()
li = [range(1,i+1)]+[range(n,i+1,-1)]+[i+1]
print(" ".join(map(str, li)))
|
s822411211
|
Accepted
| 120 | 14,052 | 258 |
n,l = map(int,input().split())
a = list(map(int,input().split()))
for i in range(n-1):
if a[i] + a[i+1] >= l:
print("Possible")
break
else:
print("Impossible")
exit()
li = list(range(1,i+1))+list(range(n-1,i+1,-1))+[i+1]
for x in li:
print(x)
|
s984681157
|
p03494
|
u792671636
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 180 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
import sys
n = input()
l = list(map(int, input().split()))
loop = 0
while (1):
for i in l:
if (i % (2 ** loop) != 0):
print(loop)
sys.exit(0)
loop = loop + 1
|
s376960301
|
Accepted
| 19 | 2,940 | 185 |
import sys
n = input()
l = list(map(int, input().split()))
loop = 0
while (1):
for i in l:
if i % (2 ** (loop + 1)) != 0:
print(loop)
sys.exit(0)
loop = loop + 1
|
s896726969
|
p03795
|
u247781603
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
price = 800
i_ate = int(input())
ans = price * i_ate - (200 * (i_ate % 15))
print(ans)
|
s391432968
|
Accepted
| 17 | 2,940 | 90 |
price = 800
i_ate = int(input())
ans = (price * i_ate) - (200 * (i_ate // 15))
print(ans)
|
s234346572
|
p02600
|
u894521144
| 2,000 | 1,048,576 |
Wrong Answer
| 35 | 9,172 | 240 |
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
def main():
X = int(input())
lst = [i * 200 + 400 for i in range(8)]
lst.reverse()
print(lst)
for i, l in enumerate(lst):
if X >= l:
print(i+1)
break
if __name__ == '__main__':
main()
|
s667599814
|
Accepted
| 29 | 9,160 | 225 |
def main():
X = int(input())
lst = [i * 200 + 400 for i in range(8)]
lst.reverse()
for i, l in enumerate(lst):
if X >= l:
print(i+1)
break
if __name__ == '__main__':
main()
|
s101166783
|
p03470
|
u721425712
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 74 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
mochi = set(map(int, input().split()))
print(len(mochi))
|
s909923229
|
Accepted
| 17 | 3,060 | 246 |
n = int(input())
d = [int(input()) for i in range(n)]
num = [0]*101
for i in range(n):
num[d[i]] += 1
ans = 0
for i in range(101):
if num[i] > 0:
ans += 1
print(ans)
|
s697257611
|
p03609
|
u365156087
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,088 | 25 |
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?
|
S = input()
print(S[::2])
|
s337936577
|
Accepted
| 29 | 9,008 | 61 |
x,t = map(int, input().split())
print(x-t if x-t >= 0 else 0)
|
s593524852
|
p03386
|
u057993957
| 2,000 | 262,144 |
Wrong Answer
| 2,223 | 2,007,064 | 130 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = list(map(int, input().split()))
x = [i for i in range(a, b+1)]
xx = list(set(x[:k] + x[-2:]))
for xi in xx:
print(xi)
|
s460835684
|
Accepted
| 18 | 3,060 | 107 |
a,b,k=map(int,input().split())
l=range(a,b+1)
ans=sorted(set(l[:k])|set(l[-k:]))
for i in ans:
print(i)
|
s866326813
|
p02534
|
u761087127
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,076 | 22 |
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
S = input()
print(S*3)
|
s051745586
|
Accepted
| 28 | 9,132 | 32 |
K = int(input())
print('ACL'*K)
|
s239598970
|
p03598
|
u923712635
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 125 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N = int(input())
K = int(input())
x = [int(x) for x in input().split()]
ans = 0
for i in x:
ans+=min(abs(K-i),i)
print(ans)
|
s055955288
|
Accepted
| 17 | 2,940 | 128 |
N = int(input())
K = int(input())
x = [int(x) for x in input().split()]
ans = 0
for i in x:
ans+=2*min(abs(K-i),i)
print(ans)
|
s070782220
|
p03469
|
u260036763
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 43 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
s.replace('7', '8', 1)
print(s)
|
s223705789
|
Accepted
| 17 | 2,940 | 33 |
s = input()
print('2018' + s[4:])
|
s593986765
|
p03759
|
u271176141
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,112 | 242 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = map(int,input().split())
if b - a == c - b:
print("Yes")
else:
print("No")
|
s166381392
|
Accepted
| 28 | 9,052 | 242 |
a,b,c = map(int,input().split())
if b - a == c - b:
print("YES")
else:
print("NO")
|
s394529369
|
p02612
|
u624455250
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,136 | 24 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input())%1000)
|
s799772144
|
Accepted
| 29 | 9,092 | 65 |
a=int(input())
r=a%1000
if r==0:
print(r)
else:
print(1000-r)
|
s852834781
|
p03545
|
u608088992
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 390 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
def Cal(A, B, op):
if op == "+":
return A + B
else:
return A - B
Num = input()
OP = ["---", "--+", "-+-", "-++", "+--", "+-+", "++-", "+++"]
for o in OP:
Sum = Cal(int(Num[0]), int(Num[1]), o[0])
Sum = Cal(Sum, int(Num[2]), o[1])
Sum = Cal(Sum, int(Num[3]), o[2])
if Sum == 7:
print(Num[0]+o[0]+Num[1]+o[1]+Num[2]+o[2]+Num[3])
break
|
s316208119
|
Accepted
| 17 | 3,064 | 616 |
import sys
def solve():
input = sys.stdin.readline
Num = list(map(int, list(input().strip("\n"))))
for i in range(8): #0: +; 1: -
ans = Num[0]
d = i
toPrint = str(Num[0])
for k in range(3):
if d % 2 == 0:
ans += Num[k + 1]
toPrint += "+" + str(Num[k + 1])
else:
ans -= Num[k + 1]
toPrint += "-" + str(Num[k + 1])
d //= 2
if ans == 7:
toPrint += "=7"
print(toPrint)
break
return 0
if __name__ == "__main__":
solve()
|
s164641345
|
p03548
|
u853900545
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 51 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z = map(int,input().split())
print(x-z//(y+1))
|
s021751309
|
Accepted
| 17 | 2,940 | 53 |
x,y,z = map(int,input().split())
print((x-z)//(y+z))
|
s632497453
|
p02972
|
u410118019
| 2,000 | 1,048,576 |
Wrong Answer
| 672 | 19,200 | 248 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
import sys
input = sys.stdin.readline
n = int(input())
a = tuple(map(int,input().split()))
b = [0] * n
for i in range(n,0,-1):
c = 0
for j in range(2,n//i):
c += b[j*i-1]
b[i-1] = (a[i-1] - c) % 2
print(sum(b))
print(' '.join(map(str,b)))
|
s461028957
|
Accepted
| 641 | 19,868 | 344 |
import sys
input = sys.stdin.readline
n = int(input())
a = tuple(map(int,input().split()))
b = [0] * n
for i in range(n,0,-1):
c = 0
for j in range(i,(n//i+1)*i,i):
c += b[j-1]
if c % 2 == a[i-1]:
continue
else:
b[i-1] = 1
d = []
for i in range(n):
if b[i] == 1:
d.append(i+1)
print(len(d))
print(' '.join(map(str,d)))
|
s813419515
|
p02262
|
u418996726
| 6,000 | 131,072 |
Wrong Answer
| 20 | 5,460 | 1 |
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
s192692976
|
Accepted
| 13,640 | 45,508 | 778 |
def insertionSort(arr,step):
count = 0
for i in range(step, len(arr)):
j = i
while j >= step and arr[j] < arr[j-step]:
count += 1
t = arr[j]
arr[j] = arr[j-step]
arr[j-step] = t
j -= step
return count
def shellSort(arr):
gaps = [776591, 345152, 153401, 68178, 30301, 13467, 5985, 2660, 1182, 525, 233, 103, 46, 20, 9, 4, 1]
m = 0
count = 0
for step in gaps:
if step > len(arr):
continue
m += 1
count += insertionSort(arr, step)
print(m)
print(" ".join(map(str, gaps[len(gaps)-m:])))
print(count)
for num in arr:
print(num)
arr = list()
for _ in range(int(input())):
arr.append(int(input()))
shellSort(arr)
|
|
s033739151
|
p02534
|
u601508807
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,136 | 23 |
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
k=int(input())
print(k)
|
s833187701
|
Accepted
| 28 | 9,152 | 101 |
k=int(input())
list=[str("ACL")]*k
ans = str()
for i in range(k):
ans = ans+str(list[i])
print(ans)
|
s406515558
|
p03973
|
u045939752
| 2,000 | 262,144 |
Wrong Answer
| 202 | 7,080 | 160 |
N people are waiting in a single line in front of the Takahashi Store. The cash on hand of the i-th person from the front of the line is a positive integer A_i. Mr. Takahashi, the shop owner, has decided on the following scheme: He picks a product, sets a positive integer P indicating its price, and shows this product to customers in order, starting from the front of the line. This step is repeated as described below. At each step, when a product is shown to a customer, if price P is equal to or less than the cash held by that customer at the time, the customer buys the product and Mr. Takahashi ends the current step. That is, the cash held by the first customer in line with cash equal to or greater than P decreases by P, and the next step begins. Mr. Takahashi can set the value of positive integer P independently at each step. He would like to sell as many products as possible. However, if a customer were to end up with 0 cash on hand after a purchase, that person would not have the fare to go home. Customers not being able to go home would be a problem for Mr. Takahashi, so he does not want anyone to end up with 0 cash. Help out Mr. Takahashi by writing a program that determines the maximum number of products he can sell, when the initial cash in possession of each customer is given.
|
N = int(input())
A = [ int(input()) for _ in range(N) ]
ans = 0
m = 0
for a in A:
if a == m+1:
m = m + 1
else:
ans += (a - 1) // (m+1)
print(ans)
|
s108219939
|
Accepted
| 213 | 7,080 | 213 |
N = int(input())
A = [ int(input()) for _ in range(N) ]
ans = 0
m = 1
for i,a in enumerate(A):
if i == 0:
ans += a-1
m = 1
elif a == m+1:
m = m + 1
else:
ans += (a - 1) // (m+1)
print(ans)
|
s165446955
|
p02690
|
u594803920
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 21,208 | 139 |
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
x = int(input())
d = {}
for a in range(-200, 200):
for b in range(-200, 200):
k = a**5 - b**5
d[k] = (a, b)
print(d[x])
|
s589382531
|
Accepted
| 162 | 22,476 | 140 |
x = int(input())
d = {}
for a in range(-200, 200):
for b in range(-200, 200):
k = a**5 - b**5
d[k] = [a, b]
print(*d[x])
|
s100360744
|
p02390
|
u445380745
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,520 | 111 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
S = int(input())
h = S / 3600
m = S % 3600 / 60
s = S % 3600 % 60
print(str(h) + ':' + str(m) + ':' + str(s))
|
s640405575
|
Accepted
| 20 | 7,592 | 113 |
S = int(input())
h = S // 3600
m = S % 3600 // 60
s = S % 3600 % 60
print(str(h) + ':' + str(m) + ':' + str(s))
|
s783174607
|
p02602
|
u933650305
| 2,000 | 1,048,576 |
Wrong Answer
| 76 | 31,492 | 222 |
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
a = list(map(int,input().split()))
b = list(map(int,input().split()))
for v in range(a[1]-a[0]):
c=list(map(numpy.prod.b[0+v:a[0]+v]))
for m in range(a[1]-a[0]):
if c[m+1]>c[m]:
print("Yes")
else:
print("No")
|
s186350454
|
Accepted
| 140 | 31,604 | 143 |
N,K=map(int,input().split())
a=list(map(int,input().split()))
for v in range(N-K):
if a[v]>=a[K+v]:
print("No")
else:
print("Yes")
|
s327620883
|
p02612
|
u169678167
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,148 | 76 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
from sys import stdin
input = stdin.readline
N = int(input())
print(N%1000)
|
s362062921
|
Accepted
| 31 | 9,152 | 132 |
from sys import stdin
input = stdin.readline
N = int(input())
if (1000- N%1000) != 1000:
print(1000- N%1000)
else:
print(0)
|
s820542224
|
p02612
|
u401086905
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,128 | 33 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print( N%100 )
|
s186459092
|
Accepted
| 32 | 9,152 | 92 |
N = int(input())
for i in range(0, 10001, 1000):
if N <= i:
break
print( i-N )
|
s364057064
|
p03836
|
u077291787
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 366 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
# ABC051C - Back and Forth
def main():
sx, sy, tx, ty = list(map(int, input().rstrip().split()))
X, Y = abs(sx - tx), abs(sy - ty)
move = ""
move += "U" * X + "R" * Y + "D" * X + "L" * Y
move += "L" + "U" * (X + 1) + "R" * (Y + 1) + "D"
move += "R" + "D" * (Y + 1) + "L" * (X + 1) + "U"
print(move)
if __name__ == "__main__":
main()
|
s735710849
|
Accepted
| 17 | 3,064 | 366 |
# ABC051C - Back and Forth
def main():
sx, sy, tx, ty = list(map(int, input().rstrip().split()))
X, Y = abs(sx - tx), abs(sy - ty)
move = ""
move += "U" * Y + "R" * X + "D" * Y + "L" * X
move += "L" + "U" * (Y + 1) + "R" * (X + 1) + "D"
move += "R" + "D" * (Y + 1) + "L" * (X + 1) + "U"
print(move)
if __name__ == "__main__":
main()
|
s787698260
|
p03090
|
u902576227
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 3,336 | 389 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n = int(input())
if n % 2 == 0:
# 1 <-> n, 2 <-> n-1... is Pair
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if i + j != n + 1:
print(str(i) + " " + str(j))
else:
# Each edge has (sum - N)
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if i + j != n:
print(str(i) + " " + str(j))
|
s147716175
|
Accepted
| 23 | 3,336 | 453 |
n = int(input())
if n % 2 == 0:
# 1 <-> n, 2 <-> n-1... is Pair
print(int(n*(n-1)/2-n/2))
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if i + j != n + 1:
print(str(i) + " " + str(j))
else:
# Each edge has (sum - N)
print(int(n*(n-1)/2-(n-1)/2))
for i in range(1, n + 1):
for j in range(i + 1, n + 1):
if i + j != n:
print(str(i) + " " + str(j))
|
s864611495
|
p03131
|
u278260569
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 277 |
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K,A,B = map(int,input().split())
if A+2 >= B:
print(1+K)
elif K <= A:
print(1+K)
else:
new_K = K-A
print(new_K)
if new_K % 2 == 1:
temp_K = int((new_K+1)/2)
print(temp_K)
print(int(temp_K*B - (temp_K-1)*A))
else:
print(int((new_K/2)*B - ((new_K/2)-1)*A + 1))
|
s387928778
|
Accepted
| 17 | 3,060 | 264 |
K,A,B = map(int,input().split())
if A+2 >= B:
print(1+K)
elif K <= A:
print(1+K)
else:
new_K = K-A
if new_K % 2 == 1:
temp_K = int((new_K+1)/2)
print((temp_K*B - (temp_K-1)*A))
else:
temp_K = int(new_K/2)
print((int(temp_K*B - (temp_K-1)*A + 1)))
|
s458805053
|
p03548
|
u428397309
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 92 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
# -*- coding: utf-8 -*-
X, Y, Z = map(int, input().split())
ans = (X + Z) // Y
print(ans)
|
s468772038
|
Accepted
| 18 | 3,064 | 98 |
# -*- coding: utf-8 -*-
X, Y, Z = map(int, input().split())
ans = (X - Z) // (Y + Z)
print(ans)
|
s306082336
|
p03474
|
u652737716
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,188 | 105 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
import re
match = re.match('^[0-9]{3}-[0-9]{4}$', input())
if match:
print('Yes')
else:
print('No')
|
s979554081
|
Accepted
| 23 | 3,188 | 129 |
import re
A, B = [x for x in input().split()]
print('Yes' if re.match('^[0-9]{' + A + '}-[0-9]{' + B + '}$', input()) else 'No')
|
s776604930
|
p03399
|
u129801138
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
a = [int(input()) for i in range(4)]
print(min(a[0:1]) + min(a[2:3]))
|
s237509603
|
Accepted
| 20 | 3,316 | 69 |
a = [int(input()) for i in range(4)]
print(min(a[0:2]) + min(a[2:4]))
|
s919346850
|
p00031
|
u197615397
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,612 | 126 |
祖母が天秤を使っています。天秤は、二つの皿の両方に同じ目方のものを載せると釣合い、そうでない場合には、重い方に傾きます。10 個の分銅の重さは、軽い順に 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g です。 祖母は、「1kg くらいまでグラム単位で量れるのよ。」と言います。「じゃあ、試しに、ここにあるジュースの重さを量ってよ」と言ってみると、祖母は左の皿にジュースを、右の皿に 8g と64g と128g の分銅を載せて釣合わせてから、「分銅の目方の合計は 200g だから、ジュースの目方は 200g ね。どう、正しいでしょう？」と答えました。 左の皿に載せる品物の重さを与えるので、天秤で与えられた重みの品物と釣合わせるときに、右の皿に載せる分銅を軽い順に出力するプログラムを作成して下さい。ただし、量るべき品物の重さは、すべての分銅の重さの合計 (=1023g) 以下とします。
|
import sys
print("\n".join((" ".join(str(2**i)for i, n in enumerate(bin(int(s))[:1:-1])if n)for s in sys.stdin.readlines())))
|
s822904141
|
Accepted
| 20 | 5,608 | 131 |
import sys
print("\n".join((" ".join(str(2**i)for i, n in enumerate(bin(int(s))[:1:-1])if n=="1")for s in sys.stdin.readlines())))
|
s437902092
|
p03853
|
u174603263
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 223 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
import sys
import math
input = sys.stdin.readline
h, w = map(int, input().rstrip().split())
pic = list(input().rstrip().split() for _ in range(h))
print(pic)
for i in range(h):
print(pic[i][0])
print(pic[i][0])
|
s731345127
|
Accepted
| 18 | 3,060 | 213 |
import sys
import math
input = sys.stdin.readline
h, w = map(int, input().rstrip().split())
pic = list(input().rstrip().split() for _ in range(h))
for i in range(h):
print(pic[i][0])
print(pic[i][0])
|
s022996922
|
p03415
|
u239301277
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 57 |
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
a=input()
b=input()
c=input()
print('a[0]'+'b[1]'+'c[2]')
|
s907083191
|
Accepted
| 17 | 2,940 | 51 |
a=input()
b=input()
c=input()
print(a[0]+b[1]+c[2])
|
s976695838
|
p02678
|
u366963613
| 2,000 | 1,048,576 |
Wrong Answer
| 2,217 | 328,928 | 738 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
# -*- coding: utf-8 -*-
import numpy as np
from collections import deque
import collections
import heapq
n, m = list(map(int, input().split()))
a = [0] * m
b = [0] * m
path = [[] for i in range(n+1)]
for i in range(m):
# a[i], b[i] = list(map(int, input().split()))
a, b = list(map(int, input().split()))
path[a].append(b)
path[b].append(a)
ans = [0]*(n+1)
visited = [1]
que = deque([[1, 1]])
now = 1
prev = 1
# BFS
while len(visited) != n:
for can_go in path[now]:
que.append([can_go, now])
now, prev = que.popleft()
if (now in visited):
continue
ans[now] = prev
visited.append(now)
for _ans in ans[2:]:
print(_ans)
|
s739125411
|
Accepted
| 858 | 61,584 | 1,009 |
# -*- coding: utf-8 -*-
import numpy as np
from collections import deque
import collections
import heapq
n, m = list(map(int, input().split()))
a = [0] * m
b = [0] * m
path = [[] for i in range(n+1)]
for i in range(m):
# a[i], b[i] = list(map(int, input().split()))
a, b = list(map(int, input().split()))
path[a].append(b)
path[b].append(a)
# start = time.time()
ans = [0]*(n+1)
visited = []
que = deque([[1, 1]])
que_items = set([1])
# BFS
while len(que) > 0:
now, prev = que.popleft()
# print('pop', now)
for can_go in path[now]:
if(not (can_go in que_items)):
# print('push', can_go)
que.append([can_go, now])
que_items.add(can_go)
ans[now] = prev
visited.append(now)
# print(visited)
if (len(visited) != n):
print('No')
else:
print('Yes')
for _ans in ans[2:]:
print(_ans)
# print(time.time() - start)
|
s344002361
|
p03739
|
u212831449
| 2,000 | 262,144 |
Wrong Answer
| 123 | 14,464 | 697 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n = int(input())
s = list(map(int,input().split()))
temp = s[0]
ans = 0
if temp > 0:
for i in range(1,n):
temp += s[i]
if i%2 == 0:
if temp <= 0:
ans += abs(temp-1)
temp = 1
else:
if temp >= 0:
ans += abs(temp+1)
temp = -1
if temp < 0:
for i in range(1,n):
temp += s[i]
if i%2 == 0:
if temp >= 0:
ans += abs(temp+1)
temp = -1
else:
if temp <= 0:
ans += abs(temp-1)
temp = 1
print(ans)
|
s641631903
|
Accepted
| 122 | 14,468 | 599 |
n = int(input())
s = list(map(int,input().split()))
temp = 0
cand1 = 0
for i in range(n):
temp += s[i]
if i%2 == 0:
if temp <= 0:
cand1 += abs(temp-1)
temp = 1
else:
if temp >= 0:
cand1 += abs(temp+1)
temp = -1
temp = 0
cand2 = 0
for i in range(n):
temp += s[i]
if i%2 == 0:
if temp >= 0:
cand2 += abs(temp+1)
temp = -1
else:
if temp <= 0:
cand2 += abs(temp-1)
temp = 1
print(min(cand2,cand1))
|
s885304864
|
p02392
|
u534156032
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,600 | 116 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
# coding: utf-8
a,b,c = [int(i) for i in input().split()]
if a < b and b < c:
print("YES")
else:
print("NO")
|
s941871911
|
Accepted
| 20 | 7,732 | 116 |
# coding: utf-8
a,b,c = [int(i) for i in input().split()]
if a < b and b < c:
print("Yes")
else:
print("No")
|
s357116021
|
p03623
|
u694370915
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 153 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
def main():
x, a, b = map(int, input().split())
if abs(a-x) > abs(b-x):
print('A')
return
print('B')
if __name__ == '__main__':
main()
|
s660320327
|
Accepted
| 17 | 2,940 | 153 |
def main():
x, a, b = map(int, input().split())
if abs(a-x) > abs(b-x):
print('B')
return
print('A')
if __name__ == '__main__':
main()
|
s449709195
|
p03470
|
u367130284
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 42 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
print(len(set(open(0).readlines()[1:-1])))
|
s431320965
|
Accepted
| 17 | 2,940 | 40 |
print(len(set(open(0).readlines()[1:])))
|
s365477467
|
p03997
|
u630027862
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 3,064 | 472 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
S_A, S_B, S_C = [list(input()) for i in range(3)]
S = S_A[0]
while True:
if S == 'a':
if len(S_A) == 0:
break
else:
S == S_A.pop(0)
elif S == 'b':
if len(S_B) == 0:
break
else:
S = S_B.pop(0)
elif S == 'c':
if len(S_C) == 0:
break
else:
S == S_C.pop(0)
if S == 'a':
print('A')
elif S == 'b':
print('B')
elif S == 'c':
print('C')
|
s343812548
|
Accepted
| 17 | 2,940 | 82 |
a, b, h = [int(input()) for i in range(3)]
S = (a + b) * h / 2
S = int(S)
print(S)
|
s304148190
|
p03377
|
u317785246
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
if a <= x and a + b >= x:
print("Yes")
else:
print("No")
|
s081282596
|
Accepted
| 17 | 2,940 | 104 |
a, b, x = map(int, input().split())
if a <= x and a + b >= x:
print("YES")
else:
print("NO")
|
s920184326
|
p02614
|
u747703115
| 1,000 | 1,048,576 |
Wrong Answer
| 150 | 9,384 | 701 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
from copy import deepcopy
h, w, k = map(int, input().split())
C = [['']*w for _ in range(h)]
for i in range(h):
C[i] = list(input())
ans = 0
for i in range(2**h-1):
for j in range(2**w-1):
_C = deepcopy(C)
H = bin(i)[2:].zfill(h)
W = bin(j)[2:].zfill(w)
for _k,_h in enumerate(H):
if _h == '1':
for _w in range(w):
_C[_k][_w] = 'x'
for _k,_w in enumerate(W):
if _w == '1':
for _h in range(h):
_C[_h][_k] = 'x'
tmp = []
_ = [tmp.extend(c) for c in _C]
if tmp.count('#')==k:
ans += 1
print(i,j,tmp,ans)
print(ans)
|
s077864469
|
Accepted
| 136 | 9,348 | 674 |
from copy import deepcopy
h, w, k = map(int, input().split())
C = [['']*w for _ in range(h)]
for i in range(h):
C[i] = list(input())
ans = 0
for i in range(2**h-1):
for j in range(2**w-1):
_C = deepcopy(C)
H = bin(i)[2:].zfill(h)
W = bin(j)[2:].zfill(w)
for _k,_h in enumerate(H):
if _h == '1':
for _w in range(w):
_C[_k][_w] = 'x'
for _k,_w in enumerate(W):
if _w == '1':
for _h in range(h):
_C[_h][_k] = 'x'
tmp = []
_ = [tmp.extend(c) for c in _C]
if tmp.count('#')==k:
ans += 1
print(ans)
|
s180311469
|
p02694
|
u161857931
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,152 | 89 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
c = 100
y = 0
while c <= X :
c = int(c * 1.01)
y += 1
print(y)
|
s872937768
|
Accepted
| 22 | 9,160 | 88 |
X = int(input())
c = 100
y = 0
while c < X :
c = int(c * 1.01)
y += 1
print(y)
|
s773417488
|
p03720
|
u146575240
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 420 |
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
# B - Counting Roads
N, M = map(int,input().split())
city = [list(map(int, input().split())) for i in range(M)]
road = [0]*N
for i in range(N):
count = 0
for j in range(M):
if i+1 == city[j][0]:
count += 1
else:
pass
if i+1 == city[j][1]:
count += 1
else:
pass
road[i] = count
print(road)
for k in range(N):
print(road[k])
|
s412041819
|
Accepted
| 18 | 3,060 | 408 |
# B - Counting Roads
N, M = map(int,input().split())
city = [list(map(int, input().split())) for i in range(M)]
road = [0]*N
for i in range(N):
count = 0
for j in range(M):
if i+1 == city[j][0]:
count += 1
else:
pass
if i+1 == city[j][1]:
count += 1
else:
pass
road[i] = count
for k in range(N):
print(road[k])
|
s008397700
|
p03759
|
u716660050
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split())
print('YES' if b-c == c-b else 'NO')
|
s267789209
|
Accepted
| 18 | 2,940 | 67 |
a,b,c=map(int,input().split())
print('YES' if b-a == c-b else 'NO')
|
s736128569
|
p03548
|
u164261323
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 48 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z = map(int,input().split())
print(x-z//y+z)
|
s504211327
|
Accepted
| 29 | 2,940 | 52 |
x,y,z = map(int,input().split())
print((x-z)//(y+z))
|
s264819308
|
p02265
|
u091533407
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,620 | 572 |
Your task is to implement a double linked list. Write a program which performs the following operations: * insert x: insert an element with key x into the front of the list. * delete x: delete the first element which has the key of x from the list. If there is not such element, you need not do anything. * deleteFirst: delete the first element from the list. * deleteLast: delete the last element from the list.
|
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 21 08:50:28 2017
@author: syaga
"""
COMMAND = ["insert", "delete", "deleteFirst", "deleteLast"]
if __name__ == "__main__":
n = int(input())
com = []
for i in range(n):
com.append(input().split())
print(com)
ans = []
for i in com:
if i[0] == COMMAND[0]:
ans.insert(0, i[1])
elif i[0] == COMMAND[1]:
ans.remove(i[1])
elif i[0] == COMMAND[2]:
ans.pop(0)
elif i[0] == COMMAND[3]:
ans.pop()
print(" ".join(ans))
|
s186000152
|
Accepted
| 1,930 | 214,316 | 601 |
# -*- coding: utf-8 -*-
from collections import deque
import sys
COMMAND = ("insert", "delete", "deleteFirst", "deleteLast")
if __name__ == "__main__":
n = int(sys.stdin.readline())
ans = deque()
inp = sys.stdin.readlines()
for i in range(n):
com = inp[i].split()
if com[0] == COMMAND[0]:
ans.appendleft(com[1])
elif com[0] == COMMAND[1]:
if com[1] in ans:
ans.remove(com[1])
elif com[0] == COMMAND[2]:
ans.popleft()
elif com[0] == COMMAND[3]:
ans.pop()
print(" ".join(ans))
|
s973852361
|
p03624
|
u996564551
| 2,000 | 262,144 |
Wrong Answer
| 40 | 9,108 | 182 |
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
S = input()
alp = list('abcdefghijklmnopqrstuvwxyz')
print(alp)
for i in range(len(S)):
if S[i] in alp:
alp.remove(S[i])
if len(alp) == 0:
print('None')
else:
print(alp[0])
|
s636435845
|
Accepted
| 39 | 9,208 | 171 |
S = input()
alp = list('abcdefghijklmnopqrstuvwxyz')
for i in range(len(S)):
if S[i] in alp:
alp.remove(S[i])
if len(alp) == 0:
print('None')
else:
print(alp[0])
|
s610634971
|
p03380
|
u867848444
| 2,000 | 262,144 |
Wrong Answer
| 104 | 14,060 | 240 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
n = int(input())
a = list(map(int,input().split()))
a = sorted(a, reverse = True)
ai = a[0] // 2
aj = 0
comb = 10 ** 10
for i in range(n):
temp = abs(a[i] - ai)
if comb >= temp:
comb = temp
aj = a[i]
print(a[0], aj)
|
s376258686
|
Accepted
| 114 | 14,428 | 239 |
n = int(input())
a = list(map(int,input().split()))
a = sorted(a, reverse = True)
ai = a[0] / 2
aj = 0
comb = 10 ** 10
for i in range(n):
temp = abs(a[i] - ai)
if comb >= temp:
comb = temp
aj = a[i]
print(a[0], aj)
|
s100161667
|
p03457
|
u350179603
| 2,000 | 262,144 |
Wrong Answer
| 341 | 3,060 | 230 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
cnt = 0
for i in range(n):
t, x, y = map(int, input().split())
if t >= x+y and t%2 == (x+y)%2:
cnt += 1
else:
print("No")
break
|
s188038465
|
Accepted
| 448 | 11,840 | 607 |
n = int(input())
cnt = 0
t = []
x = []
y = []
for i in range(n):
ttmp, xtmp, ytmp = map(int, input().split())
t.append(ttmp)
x.append(xtmp)
y.append(ytmp)
for i in range(n):
if i == 0:
if t[i] >= x[i]+y[i] and (t[i])%2 == (x[i]+y[i])%2:
cnt += 1
else:
print("No")
break
else:
if t[i] >= x[i]+y[i] and (t[i])%2 == (x[i]+y[i])%2 and x[i]+y[i]-x[i-1]-y[i-1] <= t[i]-t[i-1] :
cnt += 1
else:
print("No")
break
if cnt == n:
print("Yes")
|
s261624434
|
p02469
|
u837811962
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,660 | 238 |
Find the least common multiple (LCM) of given n integers.
|
def GCD(x,y):
r = x%y
if r == 0:
return y
else:
return GCD(y,r)
n = int(input())
list1 = list(map(int,input().split()))
result = list1[0]
for i in list1[1:]:
result = result*i/GCD(i,result)
print(result)
|
s974684925
|
Accepted
| 30 | 7,664 | 243 |
def GCD(x,y):
r = x%y
if r == 0:
return y
else:
return GCD(y,r)
n = int(input())
list1 = list(map(int,input().split()))
result = list1[0]
for i in list1[1:]:
result = result*i/GCD(i,result)
print(int(result))
|
s776944677
|
p02612
|
u697502900
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,140 | 24 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
print(int(input())%1000)
|
s757069416
|
Accepted
| 28 | 9,144 | 78 |
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000-n%1000)
|
s196828121
|
p03623
|
u383450070
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 60 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
print(min(abs(x-a),abs(x-b)))
|
s757229870
|
Accepted
| 17 | 2,940 | 71 |
x,a,b=map(int,input().split())
print("A" if abs(x-a)<abs(x-b) else "B")
|
s587915764
|
p03577
|
u504715104
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 47 |
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
x=input()
y='FESTIVAL'
print(x.rstrip(y))
|
s015261121
|
Accepted
| 19 | 2,940 | 41 |
x=input()
y='FESTIVAL'
print(x[:-8])
|
s914463378
|
p03524
|
u572373398
| 2,000 | 262,144 |
Wrong Answer
| 27 | 3,444 | 607 |
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
from collections import Counter
s = input()
c = Counter(s)
if len(c) == 1:
print('No')
elif len(c) == 2:
sort_ = sorted(c.items(), key=lambda x:x[1])
if sort_[0][1] == sort_[1][1]:
print('Yes')
else:
print('No')
else:
sort_ = sorted(c.items(), key=lambda x: x[1], reverse = True)
if sort_[0][1] - (sort_[1][1] + sort_[2][1]) > 1:
print('No')
else:
print('Yes')
|
s195677637
|
Accepted
| 25 | 3,444 | 196 |
from collections import Counter
s = input()
c = Counter(s)
x = abs(c['a'] - c['b'])
y = abs(c['a'] - c['c'])
z = abs(c['b'] - c['c'])
if max(x, y, z) < 2:
print('YES')
else:
print('NO')
|
s498061669
|
p03712
|
u778700306
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 125 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w=map(int,input().split())
print('*' * (w + 2))
for _ in range(h):
print("*" + input() + "*")
print('*' * (w + 2))
|
s434179331
|
Accepted
| 18 | 3,060 | 125 |
h,w=map(int,input().split())
print('#' * (w + 2))
for _ in range(h):
print("#" + input() + "#")
print('#' * (w + 2))
|
s071648548
|
p02743
|
u833070958
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 99 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a, b, c = map(int, input().split())
if a * b < (c - a - b) ** 2:
print("Yes")
else:
print("No")
|
s847265516
|
Accepted
| 18 | 2,940 | 118 |
a, b, c = map(int, input().split())
if c >= a + b and 4 * a * b < (c - a - b) ** 2:
print("Yes")
else:
print("No")
|
s350954803
|
p03695
|
u202570162
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 236 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N=int(input())
A=[int(i) for i in input().split()]
col=[0 for i in range(8)]
fre=0
for a in A:
if a<3200:
col[a//400]+=1
else:
fre+=1
ans=0
# print(col)
for c in col:
if c>0:
ans+=1
ans+=fre
print(ans if ans<=8 else 8)
|
s081540389
|
Accepted
| 17 | 3,064 | 264 |
N=int(input())
A=[int(i) for i in input().split()]
col=[0 for i in range(9)]
for a in A:
if a<3200:
col[a//400]+=1
else:
col[-1]+=1
ans=0
# print(col)
for i in range(8):
if col[i]>0:
ans+=1
MIN=1 if ans==0 else ans
MAX=ans+col[-1]
print(MIN,MAX)
|
s565227765
|
p04043
|
u966695411
| 2,000 | 262,144 |
Wrong Answer
| 36 | 3,064 | 90 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
L = map(int, input().split())
print("YES" if 5 in L and 7 in L and sum(L) == 17 else "NO")
|
s765521258
|
Accepted
| 40 | 3,064 | 96 |
L = list(map(int, input().split()))
print("YES" if 5 in L and 7 in L and sum(L) == 17 else "NO")
|
s616401503
|
p03369
|
u894694822
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 39 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
l=list(input())
print(700+l.count("o"))
|
s501093959
|
Accepted
| 17 | 2,940 | 46 |
l=list(input())
print(700+(l.count("o"))*100)
|
s245454366
|
p03861
|
u851648783
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 95 |
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
abx = input().split()
a, b, x = int(abx[0]), int(abx[1]), int(abx[2])
print(int(b/x - (a-1)/x))
|
s748479455
|
Accepted
| 17 | 2,940 | 102 |
abx = input().split()
a, b, x = int(abx[0]), int(abx[1]), int(abx[2])
print(int(b//x) - int((a-1)//x))
|
s802953525
|
p03069
|
u679759899
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,816 | 219 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
n = int(input())
s = input()
_min = float('inf')
for i in range(n+1):
ss = '.'*i + '#' * (n-i)
t = 0
for j in range(n):
if (s[j]) != ss[j]:
t+=1
if _min > t:
_min = t
print(t)
|
s974810751
|
Accepted
| 113 | 3,500 | 168 |
n = int(input())
s = input()
b = 0
w = 0
for i in range(n):
if s[i] == '.':
b = min(b,w) + 1
else:
b = min(b, w)
w += 1
print(min(b, w))
|
s903060144
|
p02646
|
u106095117
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,112 | 207 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
distance = abs(a - b)
distance_t = (w * t) - (v * t)
if distance + distance_t <= 0:
print('Yes')
else:
print('No')
|
s965176047
|
Accepted
| 23 | 9,204 | 207 |
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
distance = abs(a - b)
distance_t = (v * t) - (w * t)
if distance - distance_t <= 0:
print('YES')
else:
print('NO')
|
s134974494
|
p04043
|
u088488125
| 2,000 | 262,144 |
Wrong Answer
| 23 | 9,024 | 181 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a,b,c=map(int, input().split())
if a==5 and b==5 and c==7:
print("Yes")
elif a==5 and b==7 and c==5:
print("Yes")
elif a==7 and b==5 and c==5:
print("Yes")
else:
print("No")
|
s835498311
|
Accepted
| 24 | 9,036 | 181 |
a,b,c=map(int, input().split())
if a==5 and b==5 and c==7:
print("YES")
elif a==5 and b==7 and c==5:
print("YES")
elif a==7 and b==5 and c==5:
print("YES")
else:
print("NO")
|
s753249822
|
p03470
|
u033719192
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 53 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
input()
D = list(input().split())
print(len(set(D)))
|
s245797449
|
Accepted
| 17 | 3,060 | 68 |
N = int(input())
D = [input() for _ in range(N)]
print(len(set(D)))
|
s631285887
|
p03693
|
u358360701
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b = map(int, input().split())
if (g*10 + b) % 4 == 0:
print("yes")
else:
print("no")
|
s647491894
|
Accepted
| 17 | 2,940 | 102 |
r,g,b = map(int, input().split())
if (g*10 + b) % 4 == 0:
print("YES")
else:
print("NO")
|
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